Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=106 and σ=15. (a) What proportion of children aged 13 to 15 years old have scores on this test above 92 ? (Reminder: proportions are between 0 and 1 - don't put in percentages!)

Answer: 0.8238Step-by-step explanation:Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with [tex]\mu=106[/tex] and [tex]\sigma=15[/tex].Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-[tex]P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}][/tex]Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238